1. **Problem statement:** A car stopped at town C for 1 hour 40 minutes. A lorry continued its journey at 45 km/h. After the stop, the car left town C for town A and arrived at the same time as the lorry arrived at town B. We need to find the average speed of the car from town C to town A. 2. **Understanding the problem:** - The lorry travels continuously at 45 km/h. - The car stops for 1 hour 40 minutes (which is $1 + \frac{40}{60} = 1.6667$ hours). - Both arrive at their destinations simultaneously. 3. **Let:** - $d$ = distance from town C to town A (car's journey distance). - $D$ = distance from town C to town B (lorry's journey distance). - $v$ = average speed of the car from town C to town A (what we want to find). 4. **Time taken by lorry from C to B:** $$t_{lorry} = \frac{D}{45}$$ 5. **Time taken by car from C to A:** - Car stopped for 1.6667 hours, then traveled distance $d$ at speed $v$. - Total time for car: $$t_{car} = 1.6667 + \frac{d}{v}$$ 6. **Since both arrive simultaneously:** $$t_{car} = t_{lorry}$$ $$1.6667 + \frac{d}{v} = \frac{D}{45}$$ 7. **Without values for $d$ and $D$, we cannot find a numeric answer.** **Assuming the distances are equal, i.e., $d = D$, then:** $$1.6667 + \frac{d}{v} = \frac{d}{45}$$ 8. **Rearranging to solve for $v$:** $$\frac{d}{v} = \frac{d}{45} - 1.6667$$ 9. **Divide both sides by $d$ (assuming $d \neq 0$):** $$\cancel{\frac{d}{v}} = \cancel{\frac{d}{45}} - \frac{1.6667}{d}$$ $$\frac{1}{v} = \frac{1}{45} - \frac{1.6667}{d}$$ 10. **Invert to find $v$:** $$v = \frac{1}{\frac{1}{45} - \frac{1.6667}{d}}$$ **Without the distance $d$, the exact average speed cannot be determined.** --- **Summary:** To find the average speed $v$ of the car, we need the distance from town C to town A or town B. If distances are equal, the formula above applies. Otherwise, more information is required.